河北省xx中學(xué)2022-2023學(xué)年高一上學(xué)期期中數(shù)學(xué)試題（含解析）

高一上學(xué)期期中數(shù)學(xué)試題第Ⅰ卷(選擇題共60分)一、單項選擇題(本題共8小題，每小題5分，共40分，在每小題給出的四個選項中，只有一項是符合題目要求的)1.已知集合SKIPIF1<0，則SKIPIF1<0()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】【分析】由題意求出集合的交集即可.【詳解】由題意SKIPIF1<0，所以SKIPIF1<0，故選：A.【點(diǎn)睛】本題主要考查集合的交并補(bǔ)運(yùn)算，屬于簡單題.2.下列函數(shù)中，與SKIPIF1<0是同一個函數(shù)的是()A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】B【解析】【分析】根據(jù)函數(shù)的概念，結(jié)合函數(shù)的定義域與對應(yīng)法則，逐項分析即得.【詳解】對于A，函數(shù)SKIPIF1<0，與函數(shù)SKIPIF1<0的定義域不同，不是同一個函數(shù)；對于B，函數(shù)SKIPIF1<0，與函數(shù)SKIPIF1<0的定義域相同，對應(yīng)關(guān)系也相同，是同一個函數(shù)；對于C，函數(shù)SKIPIF1<0，與函數(shù)SKIPIF1<0的對應(yīng)關(guān)系不同，不是同一個函數(shù)；對于D，函數(shù)SKIPIF1<0，與函數(shù)SKIPIF1<0的定義域不同，不是同一個函數(shù).故選：B.3.命題“SKIPIF1<0有實數(shù)解”的否定是()A.SKIPIF1<0無實數(shù)解 B.SKIPIF1<0有實數(shù)解C.SKIPIF1<0有實數(shù)解 D.SKIPIF1<0無實數(shù)解【答案】D【解析】【分析】根據(jù)全稱量詞命題的否定是存在量詞命題即可求解.【詳解】因為全稱量詞命題的否定是存在量詞命題，所以“SKIPIF1<0有實數(shù)解”的否定是“SKIPIF1<0無實數(shù)解”．故選:D．4.已知函數(shù)SKIPIF1<0的對應(yīng)關(guān)系如下表所示，函數(shù)SKIPIF1<0的圖像是如圖所示的曲線SKIPIF1<0，則SKIPIF1<0的值為()x123SKIPIF1<0230A.3 B.0 C.1 D.2【答案】A【解析】【分析】根據(jù)題意，由SKIPIF1<0的圖像求出SKIPIF1<0，再由SKIPIF1<0求解即可.【詳解】根據(jù)題意，由函數(shù)SKIPIF1<0的圖像，可得SKIPIF1<0，則SKIPIF1<0故選：A．5.已知SKIPIF1<0定義域為SKIPIF1<0，則SKIPIF1<0的定義域為()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】【分析】根據(jù)復(fù)合函數(shù)的定義域求解規(guī)則求解即可.【詳解】解：因為SKIPIF1<0定義域為SKIPIF1<0，所以函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，所以，SKIPIF1<0的定義域為需滿足SKIPIF1<0，解得SKIPIF1<0.所以，SKIPIF1<0的定義域為SKIPIF1<0.故選：A6.下列說法正確的是()A.不等式SKIPIF1<0的解集為SKIPIF1<0B.若SKIPIF1<0，則函數(shù)SKIPIF1<0的最小值為2C.若實數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0D.當(dāng)SKIPIF1<0時，不等式SKIPIF1<0恒成立，則SKIPIF1<0的取值范圍是SKIPIF1<0【答案】C【解析】【分析】求出不等式的解集判斷A；由基本不等式等號成立的條件以及函數(shù)的單調(diào)性可判斷B；利用不等式的性質(zhì)可判斷C；舉反例判斷D．【詳解】對于A，不等式SKIPIF1<0的解集為SKIPIF1<0或SKIPIF1<0，故A不正確；對于B，令SKIPIF1<0，則函數(shù)SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時取等號，此時SKIPIF1<0無解，故SKIPIF1<0取不到最小值2，所以函數(shù)SKIPIF1<0的最小值不可能是2，故B錯誤，對于C，若SKIPIF1<0，則SKIPIF1<0，故C正確；對于D，當(dāng)SKIPIF1<0時，SKIPIF1<0時，不等式SKIPIF1<0恒成立，故D不正確．故選：C7.因為疫情原因，某校實行憑證入校，凡是不帶出入證者一律不準(zhǔn)進(jìn)校園，某學(xué)生早上上學(xué)，早上他騎自行車從家里出發(fā)離開家不久，發(fā)現(xiàn)出入證忘在家里了，于是回到家取上出入證，然后改為乘坐出租車以更快的速度趕往學(xué)校，令x(單位：分鐘)表示離開家的時間，y(單位：千米)表示離開家的距離，其中等待紅綠燈及在家取出入證的時間忽略不計，下列圖象中與上述事件吻合最好的是()A. B.C. D.【答案】C【解析】【分析】根據(jù)它離家的距離與離開的速度判斷．【詳解】中途回家取證件，因此中間有零點(diǎn)，排除AB，第二次離開家速度更大，直線的斜率更大，只有C滿足．故選：C．8.已知函數(shù)SKIPIF1<0，若對任意SKIPIF1<0，不等式SKIPIF1<0恒成立，則實數(shù)SKIPIF1<0的取值范圍是()A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】B【解析】【分析】先由解析式得到SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，由于SKIPIF1<0，結(jié)合SKIPIF1<0可得到SKIPIF1<0在SKIPIF1<0，SKIPIF1<0恒成立，即可得到答案．【詳解】SKIPIF1<0，因為SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，因為SKIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0，SKIPIF1<0恒成立，所以SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，所以實數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0故選：B二、多項選擇題(本題共4小題，每小題5分，共20分，在每小題給出的選項中，有多項是符合題目要求，全部選對的得5分，部分選對的得2分，有選錯的得0分)9.設(shè)函數(shù)SKIPIF1<0，當(dāng)SKIPIF1<0為增函數(shù)時，實數(shù)SKIPIF1<0的值可能是()A.2 B.SKIPIF1<0 C.SKIPIF1<0 D.1【答案】CD【解析】【分析】由題知SKIPIF1<0，且SKIPIF1<0，進(jìn)而解不等式即可得SKIPIF1<0,再結(jié)合選項即可得答案.【詳解】解：當(dāng)SKIPIF1<0時，SKIPIF1<0為增函數(shù)，則SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0為增函數(shù)，故SKIPIF1<0為增函數(shù)，則SKIPIF1<0，且SKIPIF1<0，解得SKIPIF1<0，所以，實數(shù)SKIPIF1<0的值可能是SKIPIF1<0內(nèi)的任意實數(shù).故選：CD.10.某校學(xué)習(xí)興趣小組通過研究發(fā)現(xiàn)：形如SKIPIF1<0(SKIPIF1<0，SKIPIF1<0不同時為0)的函數(shù)圖象可以由反比例函數(shù)的圖象經(jīng)過平移變換而得到，則對函數(shù)SKIPIF1<0的圖象及性質(zhì)，下列表述正確的是()A.圖象上點(diǎn)的縱坐標(biāo)不可能為1B.圖象關(guān)于點(diǎn)SKIPIF1<0成中心對稱C.圖象與x軸無交點(diǎn)D.函數(shù)在區(qū)間SKIPIF1<0上單調(diào)遞減【答案】ABD【解析】【分析】化簡SKIPIF1<0得到SKIPIF1<0，結(jié)合反比例函數(shù)SKIPIF1<0的性質(zhì)可得到結(jié)果.【詳解】SKIPIF1<0，則函數(shù)SKIPIF1<0的圖象可由的SKIPIF1<0圖象先向右平移一個單位長度，再向上平移一個單位長度得到，∴SKIPIF1<0圖象上點(diǎn)的縱坐標(biāo)不可能為1，A正確；圖象關(guān)于點(diǎn)SKIPIF1<0成中心對稱，B正確；圖象與SKIPIF1<0軸的交點(diǎn)為SKIPIF1<0，C不正確；函數(shù)在區(qū)間SKIPIF1<0上單調(diào)遞減,D正確..故選：ABD．11.已知SKIPIF1<0，SKIPIF1<0是正數(shù)，且SKIPIF1<0，下列敘述正確的是()A.SKIPIF1<0的最大值為SKIPIF1<0B.SKIPIF1<0的最小值為SKIPIF1<0C.SKIPIF1<0的最大值為SKIPIF1<0D.SKIPIF1<0的最小值為SKIPIF1<0【答案】ABD【解析】【詳解】因為SKIPIF1<0是正數(shù)，且SKIPIF1<0，所以不等式可知SKIPIF1<0，即SKIPIF1<0，得SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0取得等號，所以SKIPIF1<0的最大值為SKIPIF1<0，所以A正確;因為SKIPIF1<0是正數(shù)，且SKIPIF1<0，所以SKIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0，當(dāng)SKIPIF1<0時SKIPIF1<0有最小值為SKIPIF1<0，所以B正確;由以上知SKIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0，因為SKIPIF1<0，即SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0即SKIPIF1<0時取等號，因為SKIPIF1<0所以等號不成立，即SKIPIF1<0，所以C錯誤;因為SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0時等號成立，即SKIPIF1<0，所以SKIPIF1<0的最小值為SKIPIF1<0，所以D正確.故選:ABD.12.德國著名數(shù)學(xué)家狄利克雷(Dirichlet，1805~1859)在數(shù)學(xué)領(lǐng)域成就顯著.19世紀(jì)，狄利克雷定義了一個“奇怪的函數(shù)”SKIPIF1<0其中SKIPIF1<0為實數(shù)集，SKIPIF1<0為有理數(shù)集．則關(guān)于函數(shù)SKIPIF1<0有如下四個命題，正確的為()A.對任意SKIPIF1<0，都有SKIPIF1<0B.對任意SKIPIF1<0，都存在SKIPIF1<0，SKIPIF1<0C.若SKIPIF1<0，SKIPIF1<0，則有SKIPIF1<0D.存在三個點(diǎn)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，使SKIPIF1<0為等腰直角三角形【答案】BC【解析】【分析】根據(jù)函數(shù)的定義以及解析式，逐項判斷即可．【詳解】解：對于A選項，當(dāng)SKIPIF1<0，則SKIPIF1<0，此時SKIPIF1<0，故A選項錯誤；對于B選項，當(dāng)任意SKIPIF1<0時，存SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0；當(dāng)任意SKIPIF1<0時，存在SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0，故對任意SKIPIF1<0，都存在SKIPIF1<0，SKIPIF1<0成立，故B選項正確；對于C選項，根據(jù)題意得函數(shù)SKIPIF1<0的值域為SKIPIF1<0，當(dāng)SKIPIF1<0，SKIPIF1<0時，SKIPIF1<0，故C選項正確；對于D選項，要為等腰直角三角形，只可能為如下四種情況：①直角頂點(diǎn)SKIPIF1<0在SKIPIF1<0上，斜邊在SKIPIF1<0軸上，此時點(diǎn)SKIPIF1<0，點(diǎn)SKIPIF1<0的橫坐標(biāo)為無理數(shù)，則SKIPIF1<0中點(diǎn)的橫坐標(biāo)仍然為無理數(shù)，那么點(diǎn)SKIPIF1<0的橫坐標(biāo)也為無理數(shù)，這與點(diǎn)SKIPIF1<0的縱坐標(biāo)為1矛盾，故不成立；②直角頂點(diǎn)SKIPIF1<0在SKIPIF1<0上，斜邊不在SKIPIF1<0軸上，此時點(diǎn)SKIPIF1<0的橫坐標(biāo)為無理數(shù)，則點(diǎn)SKIPIF1<0的橫坐標(biāo)也應(yīng)為無理數(shù)，這與點(diǎn)SKIPIF1<0的縱坐標(biāo)為1矛盾，故不成立；③直角頂點(diǎn)SKIPIF1<0在SKIPIF1<0軸上，斜邊在SKIPIF1<0上，此時點(diǎn)SKIPIF1<0，點(diǎn)SKIPIF1<0的橫坐標(biāo)為有理數(shù)，則SKIPIF1<0中點(diǎn)的橫坐標(biāo)仍然為有理數(shù)，那么點(diǎn)SKIPIF1<0的橫坐標(biāo)也應(yīng)為有理數(shù)，這與點(diǎn)SKIPIF1<0的縱坐標(biāo)為0矛盾，故不成立；④直角頂點(diǎn)SKIPIF1<0在SKIPIF1<0軸上，斜邊不在SKIPIF1<0上，此時點(diǎn)SKIPIF1<0的橫坐標(biāo)為無理數(shù)，則點(diǎn)SKIPIF1<0的橫坐標(biāo)也應(yīng)為無理數(shù)，這與點(diǎn)SKIPIF1<0的縱坐標(biāo)為1矛盾，故不成立．綜上，不存在三個點(diǎn)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，使得SKIPIF1<0為等腰直角三角形，故選項D錯誤．故選：BC.【點(diǎn)睛】本題考查函數(shù)的新定義問題，考查數(shù)學(xué)推理與運(yùn)算等核心素養(yǎng)，是難題.本題D選項解題的關(guān)鍵是根據(jù)題意分直角頂點(diǎn)SKIPIF1<0在SKIPIF1<0上，斜邊在SKIPIF1<0軸上；直角頂點(diǎn)SKIPIF1<0在SKIPIF1<0上，斜邊不在SKIPIF1<0軸上；直角頂點(diǎn)SKIPIF1<0在SKIPIF1<0軸上，斜邊在SKIPIF1<0上；直角頂點(diǎn)SKIPIF1<0在SKIPIF1<0軸上，斜邊不在SKIPIF1<0上四種情況討論求解.第Ⅱ卷(共90分)三、填空題(本題共4小題，每小題5分，共20分)13.“SKIPIF1<0”是“SKIPIF1<0”的__________條件(填“充分不必要”“必要不充分”“充要”或“既不充分也不必要”)【答案】充分不必要【解析】【分析】首先解一元二次不等式，再根據(jù)充分條件、必要條件的定義判斷即可；詳解】解：由SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，因為SKIPIF1<0SKIPIF1<0，所以由SKIPIF1<0推得出SKIPIF1<0，即充分性成立；由SKIPIF1<0推不出SKIPIF1<0，即必要性不成立；所以“SKIPIF1<0”是“SKIPIF1<0”充分不必要條件；故答案為：充分不必要14.已知SKIPIF1<0，若函數(shù)SKIPIF1<0在SKIPIF1<0上SKIPIF1<0隨SKIPIF1<0增大而減小，且圖像關(guān)于SKIPIF1<0軸對稱，則SKIPIF1<0_______【答案】SKIPIF1<0【解析】【分析】利用冪函數(shù)的單調(diào)性、奇偶性與參數(shù)之間的關(guān)系可得出SKIPIF1<0的值.【詳解】若函數(shù)SKIPIF1<0在SKIPIF1<0上遞減，則SKIPIF1<0.當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0為偶函數(shù)，合乎題意；當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0為奇函數(shù)，不合乎題意.綜上所述，SKIPIF1<0.故答案為：SKIPIF1<0.15.函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上有SKIPIF1<0，則SKIPIF1<0___________.【答案】SKIPIF1<0【解析】【分析】令SKIPIF1<0，由奇偶性定義可知SKIPIF1<0為奇函數(shù)，由SKIPIF1<0可構(gòu)造方程求得結(jié)果.【詳解】令SKIPIF1<0，SKIPIF1<0，SKIPIF1<0為定義在SKIPIF1<0上的奇函數(shù)，又SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.故答案為：SKIPIF1<0.16.已知函數(shù)SKIPIF1<0為定義在SKIPIF1<0上的奇函數(shù)，滿足對SKIPIF1<0，SKIPIF1<0，其中SKIPIF1<0，都有SKIPIF1<0，且SKIPIF1<0，則不等式SKIPIF1<0的解集為________(寫成集合或區(qū)間的形式)【答案】SKIPIF1<0【解析】【分析】根據(jù)題意構(gòu)造SKIPIF1<0，判定函數(shù)SKIPIF1<0的單調(diào)性和奇偶性，利用賦值法得到SKIPIF1<0，再通過單調(diào)性和奇偶性求得不等式的解集.【詳解】解：因為SKIPIF1<0，所以當(dāng)SKIPIF1<0時，SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，又因為SKIPIF1<0為定義在R上的奇函數(shù),所以SKIPIF1<0，所以SKIPIF1<0是偶函數(shù)，且在SKIPIF1<0上單調(diào)遞減，因為SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0等價于SKIPIF1<0或SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0，即不等式SKIPIF1<0的解集為SKIPIF1<0.故答案：SKIPIF1<0.四、解答題：(本題共6小題，共70分，解答應(yīng)寫出文字說明、證明過程或演算步驟)17.已知函數(shù)SKIPIF1<0的定義域為A，集合SKIPIF1<0.(1)當(dāng)SKIPIF1<0時，求SKIPIF1<0；(2)若SKIPIF1<0，求a的取值范圍.【答案】(1)SKIPIF1<0或SKIPIF1<0(2)SKIPIF1<0【解析】【分析】(1)求出定義域，得到SKIPIF1<0，進(jìn)而計算出SKIPIF1<0及SKIPIF1<0；(2)分SKIPIF1<0與SKIPIF1<0，列出不等式，求出a的取值范圍.【小問1詳解】要使函數(shù)SKIPIF1<0有意義，則SKIPIF1<0，解得：SKIPIF1<0，所以集合SKIPIF1<0.SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0或SKIPIF1<0，∴SKIPIF1<0或SKIPIF1<0；【小問2詳解】SKIPIF1<0，①當(dāng)SKIPIF1<0時，SKIPIF1<0，即SKIPIF1<0，滿足題意；②當(dāng)SKIPIF1<0時，由SKIPIF1<0，得SKIPIF1<0，解得：SKIPIF1<0，綜上所述：a的取值范圍為SKIPIF1<0.18.已知冪函數(shù)SKIPIF1<0(實數(shù)SKIPIF1<0)的圖像關(guān)于SKIPIF1<0軸對稱，且SKIPIF1<0.(1)求SKIPIF1<0的值及函數(shù)SKIPIF1<0的解析式；(2)若SKIPIF1<0，求實數(shù)SKIPIF1<0的取值范圍.【答案】(1)SKIPIF1<0，SKIPIF1<0；(2)SKIPIF1<0.【解析】【分析】(1)由SKIPIF1<0，得到SKIPIF1<0，從而得到SKIPIF1<0，又由SKIPIF1<0，得出SKIPIF1<0的值和冪函數(shù)的解析式；(2)由已知得到SKIPIF1<0且SKIPIF1<0，由此即可求解實數(shù)SKIPIF1<0的取值范圍.【詳解】(1)由題意，函數(shù)SKIPIF1<0(實數(shù)SKIPIF1<0)的圖像關(guān)于SKIPIF1<0軸對稱，且SKIPIF1<0，所以在區(qū)間SKIPIF1<0為單調(diào)遞減函數(shù)，所以SKIPIF1<0，解得SKIPIF1<0，又由SKIPIF1<0，且函數(shù)SKIPIF1<0(實數(shù)SKIPIF1<0)的圖像關(guān)于SKIPIF1<0軸對稱，所以SKIPIF1<0為偶數(shù)，所以SKIPIF1<0，所以SKIPIF1<0.(2)因為函數(shù)SKIPIF1<0圖象關(guān)于SKIPIF1<0軸對稱，且在區(qū)間SKIPIF1<0為單調(diào)遞減函數(shù)，所以不等式SKIPIF1<0，等價于SKIPIF1<0且SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，所以實數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.【點(diǎn)睛】本題主要考查了冪函數(shù)的解析式的求解，以及冪函數(shù)的圖象與性質(zhì)的應(yīng)用，其中解答中認(rèn)真審題，熟練應(yīng)用冪函數(shù)的圖象與性質(zhì)是解答的關(guān)鍵，著重考查了推理與運(yùn)算能力，屬于基礎(chǔ)題.19.已知函數(shù)SKIPIF1<0.(1)若函數(shù)SKIPIF1<0定義域為R，求a的取值范圍；(2)若函數(shù)SKIPIF1<0值域為SKIPIF1<0，求a的取值范圍.【答案】(1)SKIPIF1<0；(2)SKIPIF1<0.【解析】【分析】(1)根據(jù)二次根式的性質(zhì)進(jìn)行求解即可；(2)根據(jù)函數(shù)值域的性質(zhì)進(jìn)行求解即可.【小問1詳解】因為函數(shù)SKIPIF1<0定義域為R，所以SKIPIF1<0在R上恒成立，當(dāng)SKIPIF1<0時，SKIPIF1<0，不符合題意；當(dāng)SKIPIF1<0時，要想SKIPIF1<0在R上恒成立，即SKIPIF1<0在R上恒成立，只需SKIPIF1<0，所以a的取值范圍為SKIPIF1<0；【小問2詳解】當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0符合題意；當(dāng)SKIPIF1<0時，要想函數(shù)SKIPIF1<0值域為SKIPIF1<0，只需SKIPIF1<0，綜上所述：a的取值范圍為SKIPIF1<0.20.已知函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的偶函數(shù)，當(dāng)SKIPIF1<0時，SKIPIF1<0是一個二次函數(shù)的一部分，其圖象如圖所示．(1)求SKIPIF1<0在SKIPIF1<0上的解析式；(2)若函數(shù)SKIPIF1<0，SKIPIF1<0，求SKIPIF1<0的最大值．【答案】(1)SKIPIF1<0(2)SKIPIF1<0【解析】【分析】(1)采用待定系數(shù)法，結(jié)合圖象可求得SKIPIF1<0在SKIPIF1<0時的解析式；由SKIPIF1<0時，SKIPIF1<0可求得SKIPIF1<0；由此可得分段函數(shù)解析式；(2)首先確定SKIPIF1<0解析式，分別在SKIPIF1<0、SKIPIF1<0和SKIPIF1<0的情況下，根據(jù)SKIPIF1<0單調(diào)性得到最大值.【小問1詳解】當(dāng)SKIPIF1<0時，結(jié)合圖象可設(shè)：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0為偶函數(shù)，SKIPIF1<0；綜上所述：SKIPIF1<0.【小問2詳解】當(dāng)SKIPIF1<0時，SKIPIF1<0，則SKIPIF1<0開口方向向下，對稱軸為SKIPIF1<0；①當(dāng)SKIPIF1<0，即SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0；②當(dāng)SKIPIF1<0，即SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0；③當(dāng)SKIPIF1<0，即SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0；綜上所述：SKIPIF1<0.21.某群體的人均通勤時間，是指單日內(nèi)該群體中成員從居住地到工作地的平均用時，某地上班族SKIPIF1<0中的成員僅以自駕或公交方式通勤，分析顯示：當(dāng)SKIPIF1<0中SKIPIF1<0(SKIPIF1<0)的成員自駕時，自駕群體的人均通勤時間為SKIPIF1<0(單位：分鐘)，而公交群體的人均通勤時間不受SKIPIF1<0影響，恒為50分鐘，試根據(jù)上述分析結(jié)果回答下列問題：(1)當(dāng)SKIPIF1<0在什么范圍內(nèi)時，公交群體的人均通勤時間少于自駕群體的人均通勤時間？(2)求該地上班族SKIPIF1<0的人均通勤時間SKIPIF1<0的表達(dá)式；并求出SKIPIF1<0的最小值.【答案】(1)SKIPIF1<0(2)SKIPIF1<0，44【解析】【分析】(1)根據(jù)題意分SKIPIF1<0、SKIPIF1<0討論，運(yùn)算求解；(2)根據(jù)題意整理求解SKIPIF1<0，結(jié)合單調(diào)性求最值.【小問1詳解】當(dāng)SKIPIF1<0時，SKIPIF1<0恒成立，公交群體的